a .685 kg block is placed on spring with a force constant k. the block is pushed downward a distance x=0.487m compressing the spring from its euilibrium length l=2.23m. the block isthen released, and the spring shoots upward. after the release, the greatest height the block reaches above the table where the spring is mounted is 4.73 m. what is the value of k?

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the answer is b

explanation:

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s